# Relation between bernoulli number recursions

The recursion formula

$$\sum_{k=0}^{n-1}{n\choose k}B_k=0$$

which is equation (34) from the MathWorld page is the most basic bernoulli number recursion.

Another recursion is given by the following:
Consider

$$\mathbf{A} = \begin{bmatrix} -\frac{1}{2} & -\frac{1}{6} & -\frac{1}{12} & -\frac{1}{20} & -\frac{1}{30} & -\frac{1}{42}\\ \frac{1}{1} & 0 & 0 & 0 & 0 & 0\\ 0 & \frac{1}{2} & 0 & 0 & 0 & 0\\ 0 & 0 & \frac{1}{3} & 0 & 0 & 0\\ 0 & 0 & 0 & \frac{1}{4} & 0 & 0\\ 0 & 0 & 0 & 0 & \frac{1}{5} & 0 \end{bmatrix}$$

This matrix was constructed by putting $-1/(1\cdot2), -1/(2\cdot3), -1/(3\cdot4), …$ on the first row and $1/1, 1/2, 1/3, …$ on the subdiagonal.

Now with $k!A^ke$ with $e:=[1, 0, 0, 0, 0, 0]^T$ and $k < 6$ we get the coefficients of the kth Bernoulli polynomial. With the constant term in the ‘first’ (leftmost) position. For example

$$3!A^3e = [0, 1/2, -3/2, 1, 0, 0]^T$$

and

$$4!A^4e = [-1/30, 0, 1, -2, 1, 0]^T$$

Together with this equation from wikipedia

$$B_n(x) = \sum_{k=0}^n{n\choose k}B_kx^{n-k}$$

we are given another bernoulli number recursion.

If these two recursions are related – in what way exactly?

#### Solutions Collecting From Web of "Relation between bernoulli number recursions"

Perhaps the thoughts which I followed recently give a satisfactory answer. I’ve looked at the “ZETA”-matrix and fiddled out the connection to the integral-representation in the Euler-MacLaurin-Formula. The article is not yet ready, needs some brushing and completing references and such. But as it might be helpful here: here is the link the integral in Euler-MacLaurin in connection with the Pascal-matrix

[update]: added the re-translation into bernoulli-numbers in the final formulae in the text